Question: A curve is defined by the parametric equations $x=8\sqrt{t}+1$ and $y=-6\sqrt{t}+t$. What is $\dfrac{d^2y}{dx^2}$ in terms of $t$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{1}{32}$ (Choice B) B $\dfrac{\sqrt{t}-3}{4}$ (Choice C) C $\dfrac{3t}{2}$ (Choice D) D $\dfrac{t-3\sqrt{t}}{16}$
Solution: We are asked to find the second derivative of a parametric function. Recall that the first derivative of a function defined parametrically by the equations $x=u(t)$ and $y=v(t)$ is found with the following rule: $\dfrac{dy}{dx}=\dfrac{\left(\dfrac{dy}{dt}\right)}{\left(\dfrac{dx}{dt}\right)}=\dfrac{v'(t)}{u'(t)}$ Then, the second derivative is found with this following rule: $\dfrac{d^2y}{dx^2}=\dfrac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{\left(\dfrac{dx}{dt}\right)}=\dfrac{\dfrac{d}{dt}\left(\dfrac{v'(t)}{u'(t)}\right)}{u'(t)}$ Let's start by finding $\dfrac{dy}{dx}$. $\dfrac{dy}{dx}=\dfrac{\sqrt{t}-3}{4}$ Now we can find $\dfrac{d^2y}{dx^2}$. $\begin{aligned} \dfrac{d^2y}{dx^2}&=\dfrac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{\left(\dfrac{dx}{dt}\right)} \\\\ &=\dfrac{\dfrac{d}{dt}\left(\dfrac{\sqrt{t}-3}{4}\right)}{\dfrac{d}{dt}(8\sqrt{t}+1)} \\\\ &=\dfrac{\left(\dfrac{t^{-\frac12}}{8}\right)}{4t^{-\frac12}} \\\\ &=\dfrac{1}{32} \end{aligned}$ In conclusion, $\dfrac{d^2y}{dx^2}=\dfrac{1}{32}$.